If \( f(x)=\begin{cases}\frac{x-2}{|x-2|}+a, & x\lt 2\\ a+b, & x=2\\ \frac{x-2}{|x-2|}+b\ , & x\gt 2\end{cases} \) is continuous at \( x=2 \), then
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( a=1,b=-1 \)
Since \( \frac{x-2}{|x-2|} \) is −1 for x < 2 and +1 for x > 2,
\( \displaystyle \lim_{x\to 2^-}f(x)=-1+a \)
\( \displaystyle \lim_{x\to 2^+}f(x)=1+b \)
\( f(2)=a+b \)
For continuity all three are equal:
\( -1+a=a+b\;\Rightarrow\;b=-1 \)
\( 1+b=a+b\;\Rightarrow\;a=1 \)